Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 3

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Theorem

Let $\left \langle {x_n} \right \rangle$ be the real sequence defined as $x_n = \paren {n + 1}^{1/n}$, using exponentiation.

Then $\left \langle {x_n} \right \rangle$ converges with a limit of $1$.


Proof

We have the definition of the power to a real number:

$\displaystyle \paren {n + 1}^{1/n} = \exp \paren {\frac 1 n \ln \paren {n+1}}$.


For $n >= 1$ then $n + 1 \le 2n$.

Hence:

\(\displaystyle \frac 1 n \ln \paren {n+1}\) \(\le\) \(\displaystyle \frac 1 n \ln \paren {2n}\) $\quad$ Logarithm is Strictly Increasing $\quad$
\(\displaystyle \) \(\) \(\displaystyle \) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 n \paren {\ln 2 + \ln n}\) $\quad$ Logarithm on Positive Real Numbers is Group Isomorphism $\quad$
\(\displaystyle \) \(\) \(\displaystyle \) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\ln 2} n + \frac 1 n \ln n\) $\quad$ $\quad$


By Powers Drown Logarithms then:

$\displaystyle \lim_{n \to \infty} \frac 1 n \ln n = 0$


By Corollary to Power of Reciprocal then:

$\displaystyle \lim_{n \to \infty} \frac 1 n = 0$


By combined sum rule for real sequences then:

$\displaystyle \lim_{n \to \infty} \paren{ \frac {\ln 2} n + \frac 1 n \ln n } = \ln 2 \cdot 0 + 0 = 0$


By Squeeze Theorem for real sequences then:

$\displaystyle \lim_{n \to \infty} \paren {n + 1}^{1/n} = 0$


Hence:

$\displaystyle \lim_{n \to \infty} \paren {n + 1}^{1/n} = \exp 0 = 1$

and the result follows.

$\blacksquare$